3. Find all points (x,y) that are 13 units away from the point (2,7) and that lie on the line x-2y=10.
We can construct a circle centered at (2,7) with a radius of 13
The equation of this circle is
(x - 2)^2 + (y - 7)^2 = 169 (1)
Rearrange the equation of the line as x = 2y+10 (2)
Put (2) int (1) for x and we have
(2y + 10 - 2)^2 + ( y - 7)^2 = 169
(2y + 8)^2 + (y - 7)^2 = 169 simplify
4y^2 + 32y + 64 + y^2 - 14y + 49 = 169
5y^2 + 18y + 113 = 169 subtract 169 from both sides
5y^2 + 18y - 56 = 0 factor this
(5y + 28) ( y - 2) = 0
Set each factor to 0 and solve for y
5y + 28 = 0 y - 2 = 0
5y = -28 y = 2
y = -28/5
Put both y values back into the equation of the line to find their associated x coordinates
x - 2(-28/5) = 10
x + 56/5 = 50/5
x = 50/5 - 56/5
x = -6/5
So one point is ( -6/5, -28/5)
And
x - 2(2) = 10
x - 4 = 10
x = 14
And the other point is (14, 2)
